Consider the following data for the next three (03) items that follow : There are 90 applicants for a job. Some of them are graduates. Some of them have less than three years experience. Let G be the event that the first applicant interviewed is a graduate and T be the event that first applicant interviewed has at least 3 years experience.Number of graduates Number of non-graduates At least 3 years experience 18 9 Less than 3 years experience 36 27
What is \(P (G \cap \overline T)\) equal to?
Calculating Probability for Job Applicants
The problem provides data about 90 job applicants, categorized by whether they are graduates and their years of experience. We are given the following data in a table format:
| At least 3 years experience (T) | Less than 3 years experience (\(\overline T\)) | Total | |
|---|---|---|---|
| Number of graduates (G) | 18 | 36 | 54 |
| Number of non-graduates (\(\overline G\)) | 9 | 27 | 36 |
| Total | 27 | 63 | 90 |
We are asked to find the probability \(P(G \cap \overline T)\). This represents the probability that the first applicant interviewed is a graduate (event G) AND has less than 3 years experience (event \(\overline T\)). The symbol \(\cap\) denotes the intersection of the events, meaning both events occur.
Understanding \(P(G \cap \overline T)\)
The term \(G \cap \overline T\) refers to the set of applicants who are both graduates and have less than 3 years of experience. Looking at the provided table:
- The row "Number of graduates (G)" gives data for graduates.
- The column "Less than 3 years experience (\(\overline T\))" gives data for those with less than 3 years experience.
- The cell where this row and column intersect shows the number of applicants who satisfy both conditions.
From the table, the number of graduates with less than 3 years experience is 36.
Calculating the Probability
The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.
In this case:
- Favorable outcomes: Applicants who are graduates AND have less than 3 years experience. Number of favorable outcomes = 36.
- Total possible outcomes: All the applicants. Total number of applicants = 90.
The probability \(P(G \cap \overline T)\) is therefore:
\(P(G \cap \overline T) = \frac{\text{Number of graduates with less than 3 years experience}}{\text{Total number of applicants}}\)
\(P(G \cap \overline T) = \frac{36}{90}\)
Simplifying the Fraction
To simplify the fraction \(\frac{36}{90}\), we can find the greatest common divisor (GCD) of 36 and 90, or simply divide by common factors:
- Divide both numerator and denominator by 2: \(\frac{36 \div 2}{90 \div 2} = \frac{18}{45}\)
- Divide both numerator and denominator by 9: \(\frac{18 \div 9}{45 \div 9} = \frac{2}{5}\)
So, \(P(G \cap \overline T) = \frac{2}{5}\).
Conclusion
The probability that the first applicant interviewed is a graduate and has less than 3 years experience is \(\frac{2}{5}\).
Revision Table: Probability of Job Applicant Characteristics
| Event | Description | Number of Applicants | Probability (Number/90) |
|---|---|---|---|
| G | Graduate | 54 | \(\frac{54}{90} = \frac{3}{5}\) |
| \(\overline G\) | Non-graduate | 36 | \(\frac{36}{90} = \frac{2}{5}\) |
| T | At least 3 years experience | 27 | \(\frac{27}{90} = \frac{3}{10}\) |
| \(\overline T\) | Less than 3 years experience | 63 | \(\frac{63}{90} = \frac{7}{10}\) |
| \(G \cap T\) | Graduate and At least 3 years experience | 18 | \(\frac{18}{90} = \frac{1}{5}\) |
| \(G \cap \overline T\) | Graduate and Less than 3 years experience | 36 | \(\frac{36}{90} = \frac{2}{5}\) |
| \(\overline G \cap T\) | Non-graduate and At least 3 years experience | 9 | \(\frac{9}{90} = \frac{1}{10}\) |
| \(\overline G \cap \overline T\) | Non-graduate and Less than 3 years experience | 27 | \(\frac{27}{90} = \frac{3}{10}\) |
Additional Information: Contingency Tables and Joint Probability
The table provided is a type of contingency table, specifically a 2x2 table, used to display the frequency distribution of two variables (in this case, Graduate Status and Experience). Such tables are very useful in calculating probabilities, especially joint probabilities and marginal probabilities.
- Joint Probability: This is the probability of two events occurring together. \(P(A \cap B)\) is a joint probability. In our case, \(P(G \cap \overline T)\) is a joint probability. The numerator comes directly from the cells within the table (like 36).
- Marginal Probability: This is the probability of a single event occurring, regardless of the outcome of the other variable. Marginal probabilities are calculated using the row or column totals (the "margins"). For example, \(P(G)\) is the total number of graduates divided by the total applicants (\(54/90\)).
- Conditional Probability: Although not asked in this question, contingency tables are also used to calculate conditional probabilities, like \(P(G | \overline T)\) (the probability of being a graduate G, given that the applicant has less than 3 years experience \(\overline T\)). This would be calculated as \(\frac{P(G \cap \overline T)}{P(\overline T)}\) or directly from the table as \(\frac{\text{Number of } G \cap \overline T}{\text{Total number of } \overline T}\), which is \(\frac{36}{63}\).
Understanding how to read and use contingency tables is fundamental for solving probability problems involving two or more categorical variables.
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